Exact solutions of equations for the Burgers

hierarchy

Nikolai A. Kudryashov ∗, Dmitry I. Sinelshchikov

Department of Applied Mathematics, Moscow Engineering and Physics Institute(State university), 31 Kashirskoe Shosse, 115409 Moscow, Russian Federation

Abstract

Some classes of the rational, periodic and solitary wave solutions for the Burgershierarchy are presented. The solutions for this hierarchy are obtained by using thegeneralized Cole - Hopf transformation.

Key words: Nonlinear evolution equations, Burgers hierarchy, Cole - Hopftransformation, exact solutions.PACS: 02.30.Jr - Ordinary differential equations

1 Introduction

The Burgers hierarchy is well known family of nonlinear evolution equations.This hierarchy can be written in the form

ut + α∂

∂x

(∂

∂x+ u

)nu = 0, n = 0, 1, 2, . . . , (1.1)

At n = 1 Eq. (1.1) is the Burgers equation

ut + 2αuux + αuxx = 0. (1.2)

Eq. (1.2) was firstly introduced in [1]. It’s well known that the Burgers equationcan be linearized by the Cole—Hopf transformation [2, 3]. Exact solutions ofEq.(1.2) were discussed in many papers( see for example [4–7] ).

∗ Corresponding author.Email address: nakudr@gmail.com (Nikolai A. Kudryashov).

1

In the case n = 2 from Eq. (1.1) we have the Sharma - Tasso - Olver (STO)equation

ut + αuxxx + 3αu2x + 3αuuxx + 3αu2 ux = 0. (1.3)

The STO equation was derived in [8,9]. Some exact solutions of this equationwas obtained in [10–16].

At n = 3 and n = 4 we have the fourth and fifth order partial differentialequations

ut + αuxxxx + 10αuxuxx + 4αuuxxx + 12αuu2x+

+6αu2uxx + 4αu3 ux = 0(1.4)

ut + αuxxxxx + 10αu2xx + 15αuxuxxx + 5αuuxxxx + 15αu3

x+

+50αuuxuxx + 10αu2uxxx + 30αu2u2x + 10αu3uxx + 5αu4ux = 0

(1.5)

In this paper we present the generalized Cole— Hopf transformation which weuse for finding different types of exact solutions: the solitary wave solutions, theperiodic solutions and the rational solutions. The advantage of our approachis that we can find the exact solutions for whole Burgers hierarchy. We canconstruct them without using the traveling wave. This fact allows us to obtainsolutions of different types.

2 Generalized Cole — Hopf transformation for solutions of theBurgers hierarchy

Eq. (1.1) can be linearized by the Cole—Hopf transformation [9, 17,18]

u =Ψx

Ψ, Ψ = Ψ(x, t) (2.1)

Taking this transformation into account, we have [18]

ut + α∂

∂x

(∂

∂x+ u

)nu =

∂

∂x

(Ψt + αΨn+1,x

Ψ

), (2.2)

where Ψn,x - n-th derivative of Ψ with respect to x.

Exact solutions of the Burgers equation can be obtained by using a general-ization of the Cole—Hopf transformation [17–23]. This transformation can bewritten as

u =FxF

+ F, F = F (x, t) (2.3)

where F (x, t) satisfies the Burgers equation. Let us show that transformation(2.3) is valid for all hierarchy (1.1). First of all, we prove the following lemma.

2

Lemma 1 The following identity takes place(∂

∂x+

Ψxx

Ψx

)nΨxx

Ψx

=Ψn+2,x

Ψx

, (2.4)

where Ψn,x is n-th derivative of Ψ with respect to x.

Proof. Let us apply the method of mathematical induction. When n = 1 weget (

∂

∂x+

Ψxx

Ψx

)Ψxx

Ψx

=Ψxxx

Ψx

(2.5)

At n = 2 we have(∂

∂x+

Ψxx

Ψx

)2Ψxx

Ψx

=

(∂

∂x+

Ψxx

Ψx

)Ψxxx

Ψx

=Ψxxxx

Ψx

(2.6)

By the induction, assuming n = k − 1, we obtain

(∂

∂x+

Ψxx

Ψx

)k−1Ψxx

Ψx

=Ψk+1,x

Ψx

(2.7)

Finally, when n = k we have

(∂

∂x+

Ψxx

Ψx

)kΨxx

Ψx

=

(∂

∂x+

Ψxx

Ψx

)(∂

∂x+

Ψxx

Ψx

)k−1Ψxx

Ψx

=

=

(∂

∂x+

Ψxx

Ψx

)Ψk+1,x

Ψx

=Ψk+2,x

Ψx

(2.8)

This equality completes the proof. 2

Theorem 1 Let F (x, t) be a solution of Eq. (1.1). Then

u =FxF

+ F (2.9)

is the solution of the Burgers hierarchy (1.1).

Proof. Using the Cole-Hopf transformation (2.1), we obtain

Ft =∂

∂x

(Ψt

Ψ

)FtF

=Ψx,t

Ψx

− Ψt

ΨFx + F 2

F=

Ψxx

Ψx

(2.10)

3

Substituting transformation (2.9) into hierarchy (1.1) and taking the Lemma1 and Eq. (2.10) into account we have following set of equalities

Fx, tF− FxFt

F 2+ Ft + α

∂

∂x

(∂

∂x+FxF

+ F

)n (FxF

+ F)

=

=∂

∂x

(FtF

)+ Ft + α

∂

∂x

(∂

∂x+Fx + F 2

F

)n (Fx + F 2

F

)=

(2.11)

=∂

∂x

(Ψx, t

Ψx

− Ψt

Ψ

)+

∂

∂x

(Ψt

Ψ

)+ α

∂

∂x

(∂

∂x+

Ψxx

Ψx

)n (Ψxx

Ψx

)=

=∂

∂x

(Ψx, t

Ψx

+ α

(∂

∂x+

Ψxx

Ψx

)n (Ψxx

Ψx

))=

(2.12)

=∂

∂x

(Ψx, t

Ψx

+ αΨn+2,x

Ψx

)=

∂

∂x

(1

Ψx

∂

∂x(Ψt + αΨn+1,x)

)= 0 (2.13)

Thus, we have that if F (x, t) satisfies equation (1.1) then u(x, t) by formula(2.9), is solution of (1.1) as well. 2

3 Solitary wave solutions of the Burgers hierarchy

Let us show that the Burgers hierarchy has the solution in the form

U(n+1, N)l (x, t) =

∑Nj=1 k

lj exp

(kj x− α kn+1

j t− x(j)0

)∑Nj=1 k

l−1j exp

(kj x− α kn+1

j t− x(j)0

) ,(j = 1, 2, ...N), (n, l = 1, 2, ...)

(3.1)

where kj and x(j)0 are arbitrary constants.

This result follows from the theorem.

Theorem. Let

U0 =ψxψ, (3.2)

be a solution of the Burgers hierarchy. Then

Uk+1 =ψk+1,x

ψk,x, ψk,x =

∂kψ

∂xk, (3.3)

4

is a solution of the Burgers hierarchy as well.

Proof. This theorem follows from the generalized transformation (2.9) for thesolution of the hierarchy (1.1). Let

U0 = F (x, t) (3.4)

be the solution of the hierarchy (1.1) equation, then

U1 =Fx + F 2

F(3.5)

is also the solution of the Burgers hierarchy by the generalized transformationfor the solution of the hierarchy (1.1).

Formulae (3.4) and (3.5) can be written in the form

U0 =ψxψ, U1 =

ψxxψx

, (3.6)

Assume that

Um =ψm,xψm−1,x

, ψm,x =∂mψ

∂xm, (3.7)

is the solution of the hierarchy (1.1) and substituting Um into the generalizedtransformation (2.9) we obtain that

Um+1 =Um,x + U2

m

Um=ψm+1,x

ψm,x(3.8)

is a solution of hierarchy (1.1).

This equality completes the proof. 2

This theorem allows us to have the solutions of the Burgers hierarchy in theform (3.1).

It is obvious, that function

ψn+1(x, t) =N∑j=0

exp ( zj), zj = kj x− α kn+1j t− x(j)

0 (3.9)

is the solution of the

ψt + αψn+1, x = 0 (3.10)

By the Cole — Hopf transformation (2.1) we have the solution of the Burgers

5

hierarchy in the form

U (n+1, N) =

∑Nj=0 kj exp ( zj),∑Nj=0 exp ( zj),

zj = kj x− α kn+1j t− x(j)

0 (3.11)

Taking the theorem into account we have the solution of the hierarchy (1.1)in the form (3.1).

Let us present some examples. When N = 2, n = l = 1 we have the solutionof the Burgers equation

U(2,2)1 =

k1 exp (z1) + k2 exp (z2)

exp (z1) + exp (z2),

zj = kjx− α k2j t− x

(j)0 , (j = 1, 2)

(3.12)

In the case N = 2, n = l = 2 we have the solution of the Sharma—Tasso—Olver equation in the form

U(3,2)2 =

k21 exp (z1) + k2

2 exp (z2)

k1 exp (z1) + k2 exp (z2),

zj = kjx− α k3j t− x

(j)0 , (j = 1, 2)

(3.13)

When N = 3, n = 2 and l = 5 we obtain the following solution of theSharma—Tasso—Olver equation

U(3,3)5 =

k51 exp (z1) + k5

2 exp (z2) + k53 exp (z3)

k41 exp (z1) + k4

2 exp (z2) + k43 exp (z3)

,

zj = kjx− α k3j t− x

(j)0 , (j = 1, 2, 3)

(3.14)

In the case N = 2, n = 3 and l = 3 we have solitary wave solution for the Eq.(1.4)

U(4,2)2 =

k31 exp (z1) + k3

2 exp (z2)

k21 exp (z1) + k2

2 exp (z2),

zj = kjx− α k4j t− x

(j)0 , (j = 1, 2)

(3.15)

We can see that Eq. (3.10) is linear and has polynomial solutions. Thus, we

6

can present its solution in the form

Ψn+1(x, t) =I∑i=0

Ci xi +

N∑j=0

ezj

zj = kjx− α kn+1j t− x(j)

0

(j = 0, 1, 2, . . . N), (n = 1, 2, . . .), N ∈ N(I = 0, 1, 2, . . . , n), (i = 0, 1, 2, . . . I)

(3.16)

From the transformation (2.1) and formula (3.16) we have the exact solutionof the hierarchy (1.1) in the form

U (n+1, N)(x, t) =

∑Ii=0 iCi x

i−1 +∑Nj=0 kje

zj∑Ii=0Ci x

i +∑Nj=0 e

zj

(j = 0, 1, 2, . . . N), (n = 1, 2, . . .), N ∈ N(I = 0, 1, 2, . . . , n), (i = 0, 1, 2, . . . I)

(3.17)

For the Burgers equation (n=1) from (3.20) we obtain the solution in the form

U (2,2)(x, t) =C1 + k1e

k1x−αk21t−x1 + k2e

k2x−αk22t−x2

C0 + C1x+ ek1x−αk21t−x1 + ek2x−αk

22t−x2

(3.18)

We demonstrate solution (3.19) when C0 = C1 = k1 = 1, k2 = 2 on Fig. 1.For the Eq. (1.4) from (3.20) we have the following solution

U (4,1)(x, t) =C1 + 2C2x+ 3C3x

2 + k1ek1x−αk4

1t−x1

C0 + C1x+ C2x2 + C3x3 + ek1x−αk41t−x1

(3.19)

-10

-5

0

5

10

x

0

5

10

t

-1

0

1

2

u

Fig. 1. The solution (3.19) of the Burgers equation

By analogy with solution (3.1) we can look for the periodic solutions of theequation for the Burgers hierarchy taking the trigonometric functions into

7

consideration. Equation (3.10) has trigonometric solutions at n + 1 = 2l + 1in the form

Ψ2l+1(x, t) =I∑i=0

Ci xi +

M∑m=0

sin z′

m +P∑p=0

cos z′

p +N∑j=0

ezj ,

zj = kj x− α k2l+1j t− xj0,

z′

m,p = k(1,2)m,p x+ (−1)l+1 α (k(1,2)

m,p )2l+1 t− x(m,p)(1,2) ,

(j = 0, 1, 2, . . . , N), (l = 1, 2, . . .), N ∈ N,(I = 0, 1, 2, . . . , 2l), (i = 0, 1, 2, . . . I),

(m, p = 0, 1, 2, . . . ,M), M, P ∈ N.

(3.20)

From the transformation (2.1) we have the exact solution of the hierarchy inthe form (1.1)

U (2l+1)(x, t) =

=

∑Ii=0 iCi x

i−1 +∑Mm=0 k

(1)m cos z

′m −

∑Pp=0 k

(2)p sin z

′p +

∑Nj=0 kje

zj∑Ii=0Ci x

i +∑Mm=0 sin z′

m +∑Pp=0 cos z′

p +∑Nj=0 e

zj

(3.21)

For example, we can write following solution for the Sharma -Tasso - Olverequation (l=1)

U (3)(x, t) =k1e

k1x−αk31t−x1 + cos (k2x+ αk3

2t− x2) k2 − sin (k3x+ αk33t− x3) k3

C0 + ek1x−αk31t−x1 + sin (k2x+ αk3

2t− x2) + cos (k3x+ αk33t− x3)(3.22)

Assuming C0 = 8, k1 = 2.2, k2 = 3, k3 = 6, x1 = 0.3, x2 = 6 and x3 = 2 wedemonstrate solution (3.23) on Fig. 2.

-5

0

5

x

0

2

4

t

-0.5

0.0

0.5

1.0

u

Fig. 2. The solution (3.23) of STO equation

8

From formula (3.21) we obtain following solution for the Eq. (1.5) (l=2)

U (5)(x, t) =C0 + k1 cos (k1x+ αk5

1t− x1)− k2 sin (k2x+ αk52t− x2)

C0 + C1x+ sin (k1x+ αk51t− x1) + cos (k2x+ αk5

2t− x2)(3.23)

Other solutions can be written using the formula (3.3).

4 Rational solutions of the Burgers hierarchy

Using Eq. (3.10) and transformations (3.3) and (3.4) we can find the rationalsolutions of the hierarchy (1.1). To obtain these solutions we use the solutionsof Eq. (3.10) in the form

ψ0(x, t) = 1, ψ1(x, t) = x, . . . , ψn(x, t) = xn (4.1)

Integrating ψn(x, t) = x with respect to x we obtain ψn+1(x, t) = x2 +ϕn+1(t).Substituting ψn+1(x, t) into Eq. (3.10) we get ϕn+1 = −(n+ 1)!α t. Substitut-ing ψn+1(x, t) into Eq. (3.10) we obtain

ψn+1(x, t) = xn+1 − (n+ 1)!α t. (4.2)

Continuing in the same way, we can obtain the solutions ψq(x, t), q = n +2, n+3, . . . as a result of integration of solution with respect to x. Taking thesepolynomial solutions of (3.10) into account we obtain the rational solutions ofthe Burgers hierarchy (1.1).

The polynomial solutions of (3.10) for n = 2 are the following

ψ0(x, t) = 1, ψ1(x, t) = x, ψ2(x, t) = x2, (4.3)

ψ3(x, t) = x3 − 6α t, (4.4)

ψ4(x, t) = x4 − 24αx t, (4.5)

ψ5(x, t) = x5 − 60αx2t, (4.6)

ψ6(x, t) = x6 − 120αx3t+ 360α2t2, (4.7)

9

ψ7(x, t) = x7 − 210αx4t+ 2520α2xt2, (4.8)

ψ8(x, t) = x8 − 336αx5t+ 10080α2x2t2, (4.9)

ψ9(x, t) = x9 − 504αx6t+ 30240α2x3t2 − 60480α3t3, (4.10)

ψ10(x, t) = x10 − 720αx7t+ 75600x4α2t2 − 604800α3xt3, (4.11)

ψ11(x, t) = x11 − 990αx8t+ 166320x5α2t2 − 3326400α3t3x2, (4.12)

ψ12(x, t) = x12 − 1320αx9t+ 332640x6α2t2−−13305600α3t3x3 + 19958400α4t4,

(4.13)

Taking into account these solutions we have the rational solutions of theSharmo—Tasso—Olver equation in the form

U1(x, t) =1

x, U2(x, t) =

2

x(4.14)

U3(x, t) = 3x2

x3 − 6α t, (4.15)

U4(x, t) = 4x3 − 6α t

x (x3 − 24α t), (4.16)

U5(x, t) = 5x3 − 24α t

x (x3 − 60α t), (4.17)

U6(x, t) = 6x2 (x3 − 60α t)

x6 − 120αx3t+ 360α2t2, (4.18)

U7(x, t) = 7x6 − 120αx3t+ 360α2t2

x (x6 − 210αx3t+ 2520α2t2), (4.19)

U8(x, t) = 8x6 − 210αx3t+ 2520α2t2

x (x6 − 336αx3t+ 10080α2t2), (4.20)

10

U9(x, t) = 9x2 (x6 − 336αx3t+ 10080α2t2)

x9 − 504αx6t+ 30240α2x3t2 − 60480α3t3, (4.21)

U10(x, t) = 10x9 − 504αx6t+ 30240α2x3t2 − 60480α3t3

x (x9 − 720αx6t+ 75600α2x3t2 − 604800α3t3), (4.22)

U11(x, t) = 11x9 − 720αx6t+ 75600α2x3t2 − 604800α3t3

x (x9 − 990αx6t+ 166320α2x3t2 − 3326400α3t3). (4.23)

-10

-5

0

5

10

x

0

2

4

t

-1

0

1

2

u

Fig. 3. The solution (4.25) of the STO equation

Using the solution of Eq.(3.10) as the sum of rational, exponential functionsand, at n = 2l, trigonometric functions we can obtain many solutions of thehierarchy (1.1). In particulare, at n = 2, taking into account solution in theform

Ψ(x, t) = C2

(1 + x2

)+ ek1 x−αk1

3t−x1 + cos(k2 x+ α k2

3t− x2

)(4.24)

we have solution of the Sharma—Tasso—Olver equation in the form

U(x, t) =2C2 x+ k1 e

k1 x−αk13t−x1 − sin(k2 x+ α k2

3t− x2

)k3

C2 (1 + x2) + ek1 x−αk13t−x1 + cos

(k2 x+ α k2

3t− x2

) (4.25)

Assuming C2 = 8, k1 = 2.2, k2 = 6, x1 = 0.3, and x2 = 2 we obtain solution(4.25) on Fig. 3.

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5 Conclusion

In this paper the generalized Cole-Hopf transformation was found for theBurgers hierarchy. We have presented some classes of the exact solutions forthe Burgers hierarchy. These classes are expressed via the rational, exponentialand triangular functions and as the sum of these functions.

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